Problem: Let $\mathbf{A}$ be a matrix such that
\[\mathbf{A} \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -15 \\ 6 \end{pmatrix}.\]Find $\mathbf{A}^5 \begin{pmatrix} 5 \\ -2 \end{pmatrix}.$
Solution: Note that
\[\mathbf{A} \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -15 \\ 6 \end{pmatrix} = -3 \begin{pmatrix} 5 \\ -2 \end{pmatrix}.\]Then
\begin{align*}
\mathbf{A}^2 \begin{pmatrix} 5 \\ -2 \end{pmatrix} &= \mathbf{A} \mathbf{A} \begin{pmatrix} 5 \\ -2 \end{pmatrix} \\
&= \mathbf{A} \left( -3 \begin{pmatrix} 5 \\ -2 \end{pmatrix} \right) \\
&= -3 \mathbf{A} \begin{pmatrix} 5 \\ -2 \end{pmatrix} \\
&= -3  \left( -3 \begin{pmatrix} 5 \\ -2 \end{pmatrix} \right) \\
&= (-3)^2 \begin{pmatrix} 5 \\ -2 \end{pmatrix}.
\end{align*}In the same way, we can compute that
\begin{align*}
\mathbf{A}^3 \begin{pmatrix} 5 \\ -2 \end{pmatrix} &= (-3)^3 \begin{pmatrix} 5 \\ -2 \end{pmatrix}, \\
\mathbf{A}^4 \begin{pmatrix} 5 \\ -2 \end{pmatrix} &= (-3)^4 \begin{pmatrix} 5 \\ -2 \end{pmatrix}, \\
\mathbf{A}^5 \begin{pmatrix} 5 \\ -2 \end{pmatrix} &= (-3)^5 \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \boxed{\begin{pmatrix} -1215 \\ 486 \end{pmatrix}}.
\end{align*}